“A1.2, step 1: a repeating key on the sculpture's tables with one, two or three of the 24 known letters carved wrong (a slip), every key length 1-26, exact, judged against scrambled copies, with planted slipped messages”

1. Take the KRYPTOS table, a key word of length 13, and write the 24 known places with their slot (place minus 1, mod 13).
2. At each known place read off the key letter that turns the known letter into the carved one.
3. Places in the same slot should give the same key letter; where two disagree, one of them is a slip.
4. Count the slips the cheapest choice needs in every slot and add them up.
5. With 3 or fewer slips the setting lives: decrypt every letter with the chosen key letters.
6. Do the same on scrambled copies of the carving, and compare how many settings live and how well they read.

key length 13, KRYPTOS table on both sides, added form: key letter forced at each known place, by slot

  slot  0: 27:O->Q key P, 66:R->P key C   slips needed 1
  slot  1: 28:R->P key C, 67:L->V key F   slips needed 1
  slot  2: 29:T->R key X, 68:I->T key P   slips needed 1
  slot  3: 30:H->N key F, 69:N->T key L   slips needed 1
  slot  4: 31:E->G key C, 70:C->M key J   slips needed 1
  slot  5: 32:A->K key T, 71:L->Z key I   slips needed 1
  slot  6: 33:S->S key A, 72:O->F key H   slips needed 1
  slot  7: 34:T->S key C, 73:C->P key U   slips needed 1
  slot  8: 22:E->F key B, 74:K->K key A   slips needed 1
  slot  9: 23:A->L key K   slips needed 0
  slot 10: 24:S->R key V   slips needed 0
  slot 11: 25:T->V key S, 64:B->N key L   slips needed 1
  slot 12: 26:N->Q key B, 65:E->Y key R   slips needed 1

slips needed at length 13: 11 (more than 3: dead)

A1 (knowledge/12-assumptions.md): every exact verdict assumed no enciphering slip; K1-K3 carry one enciphering slip (K2's dropped S) and the PALIMPCEST worksheet shows Sanborn working one letter at a time against a written-out key (C27(a), 01-facts.md). Sanborn 2005: 'Saying I was sloppy is a good way of saying it' (02-sanborn-said.md). Scheidt 2015: if Jim made an encoding error he would not have known (03-scheidt-said.md). C1(c) and C27(c) for the table; C17(a)/(b) for a copper not as intended. Model: c_i = C[P(p_i) + k[i mod L]] or C[k[i mod L] - P(p_i)] with P, C each the standard or KRYPTOS alphabet (8 classes; variant Beaufort is the added form with the key negated; the key alphabet only relabels key values). A slip at known place q: the carved letter is not the one the key gives. For key length L each known place forces a key value in its slot (q - 1 mod L); a setting is a choice of value per slot among the values its known letters force; its number of slips is the number of known letters that disagree with their slot's chosen value. A slip at a known place alone in its slot fails no check: it garbles one column of the reading and is A1.0's case (the judge reads through it). Statistic: settings with exactly d = 0-3 failed checks per length, K4 against each copy; and the blind judge's best score on each text's readings. Cost of naming the slips: 4.7 k + log2 C(24, k) bits (9.3, 17.5, 25.1 for k = 1, 2, 3). Key lengths 27-29 and 53+ have no check at all on these tables, so no slip can show there and they are not rerun (F1/2, 0034, 0048, 0053).

On K4 no setting with 3 or fewer slips exists at key lengths 1-23 (at least 7 checks there, so 4 or more must hold by chance); at 24-26 settings exist on K4 and on every copy in about equal numbers, because 1-5 checks are all there is. No K4 reading clears the quadgram cut or scores above every copy's best with the judge. The three planted messages (key lengths 13, 20 and 25, with 2, 3 and 3 slips) are found, each its own text's top reading by the judge and above all of its own copies. Zero candidates pass gate 4.

1. For each text, key length L 1-26 and class, the forced key value at each known place.
2. Per slot: its distinct values, each with the number of its known letters that would then be slips.
3. Enumerate every setting with at most 3 slips (depth first over the slots).
4. Count by number of slips; K4 against 20 copies.
5. Each setting's reading ('?' where no known letter fixes the slot), quadgram score of the letters outside the known places.
6. For every text, its best reading and every reading at or above -5.0 go to the judge, K4's, the copies', the plants' and the plants' copies' in one call.
7. A text's reading counts only above every one of its own copies' best.

All 50 candidates, lowest loss first.